I'm trying to figure out how to perform a successive XOR on a row in matlab, where each element is the result of XORing itself and the previous element, for example:
If the row is
x = [1 0 1 1]
I would want the result to be:
x[0] = 1
x[1] = x[0]^x[1] = 1
x[2] = x[1]^x[2] = 0
x[3] = x[2]^x[3] = 1
x = [1 1 0 1]
I have tried using xor(A,B) but this only seems to work on multiple arrays at once. I also have tried this loop:
for k = 10000:1
for i = 1:64
allchallenge_full(k,i+1) = xor(allchallenge_full(k,i).^allchallenge_full(k,i+1))
end
end
But this simply results in all 1s.
Any suggestions are appreciated! Thank you!
If the input x is just zeros and ones:
result = mod(cumsum(x), 2);
To apply it on each row of a matrix x:
result = mod(cumsum(x,2), 2);
If you want to keep things simple, a for loop, together with the xor function (^ in Matlab represents the raising to a power), should work fine:
x = [1 0 1 1];
for i = 2:numel(x)
x(i) = xor(x(i-1),x(i));
end
The final output is the expected one:
x =
1 1 0 1
Remember that, in Matlab, indexing is one-based and not zero-based as in many other programming languages. The first element of x is, therefore, x(1) and not x(0).
I would like to implement code which will add two integers by their binary representation.
For instance:
a = [1 1 1 0];
and
b = [1 0 1 1];
So I implemented the following algorithm:
function s=add_binary(a,b)
m=length(a);
s=zeros(m+1,0);
n=length(a);
c=0;
for ii=n:-1:1
d=floor((a(ii)+b(ii)+c)/2);
s(ii)=a(ii)+b(ii)+c-2*d;
c=d;
end
s(1)=c;
end
However, it returns this as a result:
s=add_binary(a,b)
s =
1 0 0 1
Here, there should be an additional 1 on the left, such that it would read:
1 1 0 0 1
Where am I making a mistake?
Since you're dealing with binary numbers, why not use logical operators?
function s = add_binary(a, b)
a = [0 a];
b = [0 b];
c = a&b;
while any(c)
b = xor(a, b);
a = circshift(c, -1);
c = a&b;
end
s = a+b;
end
And a test:
>> a = randi([0 1], [1 6])
a =
0 1 0 1 0 1
>> b = randi([0 1], [1 6])
b =
1 1 0 0 0 1
>> s = add_binary(a, b)
s =
1 0 0 0 1 1 0
here is my solution
function s=add_binary(a,b)
m=length(a);
s=zeros(m+1,1);
n=length(a);
c=0;
for ii=n:-1:1
d=floor((a(ii)+b(ii)+c)/2);
s(ii+1)=a(ii)+b(ii)+c-2*d;
c=d;
end
s(1)=c;
s=s';
end
>> s=add_binary(a,b)
s =
1 1 0 0 1
This problem was discussed in a 2011 question on MATLAB Answers. I'd like to mention 2 more solutions:
If you have the Communications System Toolbox, you can use bi2de and de2bi:
de2bi(bi2de(a,'left-msb') + bi2de(b,'left-msb'),'left-msb')
If the amount of "digits" in the input is known (and reasonable), one solution could be to store all possible input combinations in some data-structure, which would facilitate rapid access. One such example is the MapN class (though it can just as easily be done with regular numeric arrays). I will provide a small example to show this idea (including a benchmark):
function allResults = q43095156(allResults)
if nargin == 0
% Let's store all combinations of 8 binary digits:
in = uint8(0:255);
out = uint16(in)+uint16(in.');
outB = logical(de2bi(out(:),'left-msb'));
[in1,in2] = meshgrid(in,in);
allResults = MapN(...
num2cell([num2cell(in1(:),2),num2cell(in2(:),2)],2),...
num2cell(outB,2));
end
benchmark(allResults);
end
function benchmark(allResults)
rng(43095156);
a = logical(randi([0 1],10,8,'uint8'));
b = logical(randi([0 1],10,8,'uint8'));
% Test:
R{5} = de2bi(bi2de(a,'left-msb') + bi2de(b,'left-msb'),'left-msb');
R{4} = add_a_bunch_gnovice(a,b);
R{3} = add_a_bunch_Sardar(a,b);
R{2} = add_a_bunch_dato(a,b);
R{1} = getFromMap(a, b, allResults);
assert(isequal(R{:}));
% Benchmark:
a = logical(randi([0 1],1000,8,'uint8'));
b = logical(randi([0 1],1000,8,'uint8'));
fprintf(1,'\nSardar''s method:\t%f',timeit(#() add_a_bunch_Sardar(a, b) ));
fprintf(1,'\nDev-iL''s method:\t%f',timeit(#() getFromMap(a, b, allResults) ));
fprintf(1,'\ngnovice''s method:\t%f',timeit(#() add_a_bunch_gnovice(a, b) ));
fprintf(1,'\ndato''s method:\t\t%f',timeit(#() add_a_bunch_dato(a, b) ));
fprintf(1,'\nbi2de method:\t\t%f',timeit(#() de2bi(bi2de(a,'left-msb') + bi2de(b,'left-msb'),'left-msb')));
end
function out = getFromMap(a,b,map)
out = cell2mat(values(map, num2cell([ num2cell(bi2de(a,'left-msb'),2),...
num2cell(bi2de(b,'left-msb'),2)],2)));
end
function out = add_a_bunch_gnovice(a,b)
out = zeros(size(a)+[0,1],'logical');
for ind1 = 1:size(a,1)
out(ind1,:) = add_binary_gnovice(a(ind1,:), b(ind1,:));
end
end
function out = add_a_bunch_Sardar(a,b)
out = zeros(size(a)+[0,1],'logical');
for ind1 = 1:size(a,1)
out(ind1,:) = add_binary_Sardar(a(ind1,:), b(ind1,:));
end
end
function out = add_a_bunch_dato(a,b)
out = zeros(size(a)+[0,1],'logical');
for ind1 = 1:size(a,1)
out(ind1,:) = add_binary_dato(a(ind1,:), b(ind1,:));
end
end
function s = add_binary_gnovice(a, b)
a = [0 a];
b = [0 b];
c = a&b;
while any(c)
b = xor(a, b);
a = circshift(c, -1);
c = a&b;
end
s = a+b;
end
function s = add_binary_Sardar(a,b)
s = logical(str2double(num2cell(dec2bin(bin2dec(num2str(a)) +...
bin2dec(num2str(b)), numel(a)+1))));
end
function s = add_binary_dato(a,b)
m = length(a);
s = zeros(m+1,1);
n = length(a);
c = 0;
for ii = n:-1:1
d = floor((a(ii)+b(ii)+c)/2);
s(ii+1) = a(ii)+b(ii)+c-2*d;
c = d;
end
s(1) = c;
s = logical(s.');
end
And the results with my x64 MATLAB R2017a # Win10 are:
Sardar's method: 0.336414
Dev-iL's method: 0.061656
gnovice's method: 0.022031
dato's method: 0.002123
bi2de method: 0.000356
So this should give us an idea which methods are faster (unless I messed the wrapper functions up greatly)...
The advantage of the MapN method would be noticeable if the computation of the sum was some other, costlier operation.
Use num2str to convert a and b to strings and use bin2dec to convert them to decimal. Then add them and convert the sum back to binary using dec2bin. Use num2cell to split the string and finally use str2double to get the desired result.
s=str2double(num2cell(dec2bin(bin2dec(num2str(a))+bin2dec(num2str(b)))))
% Output for given a and b
%---------------------------
% s =
% 1 1 0 0 1
Hi everyone this is What I did to carry out an iteration method(gauss seidel) and I want when iteration number greater than 30 it will stop and generate the corresponding result up to 30 iteration. But I wonder why the output result were so weird and I try to check the value on the command window by typing x_ans(:,1) it gives me the correct value. It really made me frustrated why the generate result were not the same. Or any other circumstance or function can be used to set for not converging condition. Sincerely thanks in advance for every single help.
clear;clc
A = [2 8 3 1;0 2 -1 4;7 -2 1 2;-1 0 5 2]
B = [-2;4;3;5]
Es = 1e-5
n = length(B);
x = zeros(n,1);
Ea = ones(n,1);
iter = 0;
while max(Ea) >= Es
if iter <= 30
iter = iter + 1;
x_old = x;
for i = 1:n
j = 1:n;
j(i) = [];
x_cal = x;
x_cal(i) = [];
x(i) = (B(i) - sum(A(i,j) * x_cal)) / A(i,i);
end
else
break
end
x_ans(:,iter) = x;
Ea(:,iter) =abs(( x - x_old) ./ x);
end
result = [1:iter; x_ans; Ea]'
I've gone through the formulas and they are all OK. On a side note, the sum is not necessary. The problem lies with your input data - try reordering! check for example the following, which works
A = [7 -2 1 2;
2 8 3 1;
-1 0 5 2;
0 2 -1 4;]
B = [3;-2;5;4]
see the wiki under convergence.
If I have a matrix and I want to apply a function to each row of the matrix. This function has three possible outputs, either x = 0, x = 1, or x > 0. There's a couple things I'm running into trouble with...
1) The cases that output x = 1 or x > 0 are different and I'm not sure how to differentiate between the two when writing my script.
2) My function isn't counting correctly? I think this might be a problem with how I have my loop set up?
This is what I've come up with. Logically, I feel like this should work (except for the hiccup w/ the first problem I've stated)
[m n] = size(matrix);
a = 0; b = 0; c = 0;
for i = 1 : m
x(i) = function(matrix(m,:));
if x > 0
a = a + 1;
end
if x == 0
b = b + 1;
end
if x == 1
c = c + 1;
end
end
First you probably have an error in line 4. It probably should be i instead of m.
x(i) = function(matrix(i,:));
You can calculate a, b and c out of the loop:
a = sum(x>0);
b = sum(x==0);
c = sum(x==1);
If you want to distinguish x==1 and x>0 then may be with sum(xor(x==1,x>0)).
Also you may have problem with precision error when comparing double values with 0 and 1.
Let me describe my problem with an example. Suppose we have matrix A:
A =
1 0 1
1 1 1
0 1 1
and matrix B:
B =
1 1
1 1
How do i write function C = func(A, B) to check if B exists in A or not?
If it exists in A, the function returns C = [0 0 0; 0 1 1; 0 1 1], and if it does not, the function returns C = [0 0 0; 0 0 0; 0 0 0];.
Edit:
It should be mentioned that if A is m-by-n, and B is p-by-q, then m > p and p > q always.
Thanks in advance.
The most efficient requires the signal processing toolbox. Then you can simply use xcorr2(). Following your example, the following should work:
C = xcorr2 (A, B);
[Row, Col] = find (C == max (C(:)));
%% these are not the coordinates for the center of the best match, you will
%% have to find where those really are
%% The best way to understand this is to strip the "padding"
row_shift = (size (B, 1) - 1)/2;
col_shift = (size (B, 2) - 1)/2;
C = C(1+row_shift:end-row_shift, 1+col_shift:end-col_shift)
[Row, Col] = find (C == max (C(:)));
if (B == A(Row-row_shift:Row+row_shift, Col-col_shift:Col+col_shift))
disp ("B shows up in A");
endif
The code above looks a bit convoluted because I'm trying to cover inputs of any size (still only odd sized) but should work.
If you don't have this toolbox, I think you can use the Octave code for it which should require only small adjustments. Basically, the only three lines that matter follow (note that the code is under GPL):
[ma,na] = size(a);
[mb,nb] = size(b);
c = conv2 (a, conj (b (mb:-1:1, nb:-1:1)));
In Octave, if you are using at least the signal package 1.2.0, xcorr2 can also take an extra option "coeff" which calculates the normalized cross correlation. This will have a value of 1 when the match is perfect, so you can simplify this with:
C = xcorr2 (A, B, "coeff");
if (any (C(:) == 1)
display ("B shows up in A");
endif
I would do a loop to check resized sub-matrix of A. My code stops after B is found once but it could be updated to count the number of found occurrences.
A = [1 0 1; 1 1 1; 0 1 1; 0 0 0];
B = [1 1; 1 1];
[m n] = size(A);
[p q] = size(B);
found = 0;
x = 0;
while ~found && (x+p-1 < m)
y = 0;
x = x + 1;
while ~found && (y+q-1 < n)
y = y + 1;
A(x:x+p-1,y:y+q-1)
found = isequal(A(x:x+p-1,y:y+q-1),B);
end
end
fprintf('Found Matrix B in A(%d:%d, %d:%d)\n',x,x+p-1,y,y+q-1);
Maybe there is some vectored way, but this is a straight forward function which just slides across A and checks for B:
function ret=searchmat(A,B)
ret=zeros(size(A));
for k=0:size(A,1)-size(B,1)
for l=0:size(A,2)-size(B,2)
if isequal(A(1+k:size(B,1)+k,1+l:size(B,2)+l),B)
ret(1+k:size(B,1)+k,1+l:size(B,2)+l)=1; % or B as you wish
end
end
end
end
If B is discovered more than once, it's also marked with ones. If you want to quit after the first match, you can just put in a return statement in the if clause.